HW 4.5.1.b: Double Angle Formulas

1. If sin x( ) = 14 and x is in quadrant I, then find exact values for (without solving for x):

a. ( )sin 2x b. ( )cos 2x c. ( )tan 2x

2. If cos x( ) = 35 and x is in quadrant I, then find exact values for (without solving for x):

a. ( )sin 2x b. ( )cos 2x c. ( )tan 2x

Evaluate each expression without a calculator. 3. cos2 22.5°( )− sin2(22.5°) 4. 2cos2 67.5°( )−1

5. 2sin 112.5!( )cos 112.5!( ) 6.

2 tan 75!( )1− tan2 75!( )

Simplify each expression. 7. ( )2 2cos 9 sin (9 )x x− 8. ( )2 2cos 6 sin (6 )x x− 9. ( )4sin 8 cos(8 )x x 10. ( )6sin 5 cos(5 )x x

Solve for all solutions on the interval [0, 2 )π . 11. 6sin 2t( ) + 6sin t( ) = 0 12. sin 2t( ) + 3cos t( ) = 0 13. cos 2θ( ) = 6cos2 θ( )− 4 14. 22cos 2cos2 1x x− = 15. ( ) ( )sin 2 cost t= 16. ( ) ( )cos 2 sint t= 17. cos 6x( )− cos 3x( ) = 0 18. sin 2x( )sin x( ) = cos x( ) Prove the identity, Manipulate the left side to equal the right side. 19. ( ) ( )2sin cos 1 sin 2t t t− = − 20. ( ) ( )22 4sin 1 cos 2 sinx x x− = +

21. cos4 x − sin4 x = cos 2x( ) 22. ( )( ) ( )sin 2

tan1 cos 2

θθ

θ=

+

Selected Answers: 1. a. sin(2𝑥) = 2 sin 𝑥 cos 𝑥

To find cos 𝑥:

sin! 𝑥 + cos! 𝑥 = 1

cos! 𝑥 = 1− !!"= !"

!"

cos 𝑥 = ± !"!"

Note that we need the positive root since we are told 𝑥 is in quadrant 1.

cos 𝑥 = !"!

So: sin(2𝑥) = 2 !!

!"!

= !"!

b. cos(2𝑥) = 2 cos! 𝑥 − 1

= (2 !"!")− 1 = !"!!"

!"= !

!

c. tan(2𝑥) = !"#(!!)!"#(!!)

= !"!

!!= !"

!

3. cos! 𝑥 – sin! 𝑥 = cos 2𝑥 , so cos!(22.5°) – sin!(22.5°) = cos 45° = √!!

5. 2sin 𝑥 cos 𝑥 = sin(2𝑥), so 2sin 112.5° cos 112.5° = sin 225° = − √!!

7. cos!(9𝑥) – sin!(9𝑥) = cos 2 9𝑥 = cos(18𝑥)

9. 4 sin(8𝑥) cos(8𝑥) = 2 (2sin(8𝑥) cos(8𝑥)) = 2 sin(16𝑥)

11. 6 sin(2𝑡) + 6 sin 𝑡 = 6 ∙ 2 sin 𝑡 cos 𝑡 + 6 sin 𝑡 = 6 sin 𝑡 (2 cos 𝑡 + 1), so we can solve

6sin 𝑡 = 0 or cos 𝑡 = −1/2

𝑡 = 0, 𝜋 or 𝑡 = !!!, !!!

13. cos 2𝜃 = 6 cos! 𝜃 − 4, 2 cos! 𝜃 − 1 = 6 cos! 𝜃 − 4

3 = 4 cos! 𝜃, ± 3/2 = cos𝜃

𝜃 = !!, !!!, !!!, !!!!

15. sin 2𝑡 = cos 𝑡

2 sin 𝑡 cos 𝑡 = cos 𝑡

2 sin 𝑡 cos 𝑡 − cos 𝑡 = 0

cos 𝑡 2 sin 𝑡 − 1 = 0

cos 𝑡 = 0 or 2 sin 𝑡 − 1 = 0

If cos 𝑡 = 0, then 𝑡 = !! or !!

!. If 2 sin 𝑡 − 1 = 0, then sin 𝑡 = !

!, so 𝑡 = !

! or !!

!. So 𝑡 = !

!, !!

!, !! or !!

!.

17. cos 6𝑥 − cos 3𝑥 = 0

2 cos! 3𝑥 − 1− cos(3𝑥) = 0

2 cos! 3𝑥 − cos(3𝑥)− 1 = 0

(2 cos(3𝑥) + 1)(cos(3𝑥))− 1 = 0

cos 3𝑥 = − !! or 1

Since we need solutions for 𝑥 in the interval 0, 2𝜋 , we will look for all solutions for 3𝑥 in the interval

[0, 6𝜋). If cos 3𝑥 = − !!, then there are two possible sets of solutions. First, 3𝑥 = !!

!+ 2𝜋𝑘 where

𝑘 = 0, 1, or 2, so 𝑥 = !!!+ !!"

! where 𝑘 = 0, 1, or 2. Second,

3𝑥 = !!!+ 2𝜋𝑘 where 𝑘 = 0, 1, or 2, so 𝑥 = !!

!+ !!"

! where 𝑘 = 0, 1, or 2 If cos 3𝑥 = 1, then

3𝑥 = 2𝜋𝑘 where 𝑘 = 0, 1, or 2, so 𝑥 = !!"!

where 𝑘 = 0, 1, or 2.